# How do you calculate the mass percent composition of carbon in each carbon-containing compound?

## a. $C {H}_{4}$ b. ${C}_{2} {H}_{6}$ c. ${C}_{2} {H}_{2}$ d. ${C}_{2} {H}_{5} C l$

Oct 15, 2016

The percentages by mass are a. = 74.91; b. = 79.89; c. 92.26; d. 37.23.

#### Explanation:

The method is the same for each compound.

The formula for mass percent is

color(blue)(bar(ul(|color(white)(a/a) "Mass %" = "mass of element"/"mass of compound" × 100 %color(white)(a/a)|)))" "

a. ${\text{CH}}_{4}$

$\text{Mass of 1 molecule of CH"_4 = "12.01 u + 4×1.008 u" = "16.032 u}$

$\text{Mass of C" = "12.01 u}$

$\text{% by mass of C "= (12.01 color(red)(cancel(color(black)("u"))))/(16.032 color(red)(cancel(color(black)("u")))) × 100 % = "74.91 %}$

b. ${\text{C"_2"H}}_{6}$

$\text{Mass of 1 molecule of C"_2"H"_6 = "2×12.01 u + 6×1.008 u" = "24.02 u + 6.048 u" = "30.068 u}$

$\text{Mass of C" = "24.02 u}$

$\text{% by mass of C " = (24.02 color(red)(cancel(color(black)("u"))))/(30.068 color(red)(cancel(color(black)("u")))) × 100 % = "79.89 %}$

c. ${\text{C"_2"H}}_{2}$

$\text{Mass of 1 molecule of C"_2"H"_2 = "2×12.01 u + 2×1.008 u" = "24.02 u + 2.016 u" = "24.036 u}$

$\text{Mass of C" = "24.02 u}$

$\text{% by mass of C " = (24.02 color(red)(cancel(color(black)("u"))))/(26.036 color(red)(cancel(color(black)("u")))) × 100 % = "92.26 %}$

d. $\text{C"_2"H"_5"Cl}$

$\text{Mass of 1 molecule of C"_2"H"_5"Cl" = "2×12.01 u + 5×1.008 u + 35.45 u" = "24.02 u + 2.016 u + 35.45 u" = "64.51 u}$

$\text{Mass of C" = "24.02 u}$

$\text{% by mass of C " = (24.02 color(red)(cancel(color(black)("u"))))/(64.51 color(red)(cancel(color(black)("u")))) × 100 % = "37.23 %}$